Asymptotic forms of Legendre functions for large degree

Question

Does anyone know where to find (or how to obtain) expressions for the Legendre functions for large degree, to second order? For example, to first order the expressions are $$ P_n(\cosh(x)) ~ \substack{\huge\rightarrow\\\scriptstyle{n\rightarrow\infty}}~ \frac{1}{\sqrt{\pi n}}\frac{e^{(n+1/2)x}}{\sqrt{2\sinh(x)}} \\ Q_n(\cosh(x))~ \substack{\huge\rightarrow\\\scriptstyle{n\rightarrow\infty}}~ \sqrt{\frac{\pi}{n}}\frac{e^{-(n+1/2)x}}{\sqrt{2\sinh(x)}}, $$ for integer $n$. These are apparently given on page 305-306 of "The theory of spherical and ellipsoidal harmonics" by E. W. Hobson, although I can't seem to access this book.

I am looking for the second order of these expressions - corrections $\mathcal{O}(1/n)$ relative to the above expressions.

Answer 1

A series with precise error estimates is derived in Error bounds for a uniform asymptotic expansion of the Legendre function:

$$P_n(\cosh x)=\left(\frac{x}{\sinh x}\right)^{1/2}\sum_{\nu=0}^{\infty}c_\nu(x)\frac{I_\nu[(n+1/2)x]}{(n+1/2)^\nu},$$ $$c_0=1,\;\;c_1=\frac{1}{8}(\coth x-1/x),$$ $$c_2=-\frac{1}{16}(1+3/x^2-(3/x)\coth x)+\frac{9}{128}x^{-2}(1-x\coth x)^2.$$

(There is a recursion relation for higher order $c_\nu$'s.)
An alternative representation involving only the Bessel function $I_0$ is given in equation 5.1 of that paper.

Answer 2

In addition to the Carlo Beenakker's answer. The following asymptotic expansion was proved in https://www.sciencedirect.com/science/article/pii/0041555365901345?via%3Dihub (Asymptotic formulae for legendre functions, by N.K.Chukhrukidze. Russion version can be downloaded from here http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=zvmmf&paperid=7560&option_lang=eng): $$P_n(\cosh{x})=A_0(x)\left[ I_0[(n+1/2)x]\left(1+\frac{A_1(x)}{(n+1/2)^2} -\frac{4A_3(x)-xA_4(x)}{(n+1/2)^4x}\\+\frac{24A_4(x)}{(n+1/2)^6x^2}+\ldots\right)+I_1[(n+1/2)x]\left(\frac{A_1(x)}{n+1/2}-\frac{2A_2(x)-xA_3(x)}{(n+1/2)^3x}+\\\frac{8[A_3(x)-xA_4(x)]}{(n+1/2)^5x^2}-\frac{48A_4(x)}{(n+1/2)^7x^3}+\ldots \right)\right],$$ where $$A_0(x)=\sqrt{\frac{x}{\sinh{x}}},\;\;A_1(x)=\frac{x\coth{x}-1}{8x},\\A_2(x)=\frac{9x^2\coth^2{x}+6x\coth{x}-8x^2-15}{128x^2},\\A_3(x)=\frac{5\left(15x^3\coth^3{x}+27x^2\coth^2{x}-11x^3\coth{x}+21x\coth{x}-24x^2-63\right)}{1024x^3},A_4(x)=\frac{7}{32768x^4}\left(525x^4\coth^4{x}+1500x^3\coth^3{x}-720x^4\coth^2{x}\\+2430x^2\coth^2{x}-1600x^3\coth{x}+1980x\coth{x}+192x^4-2160x^2-6435\right).$$

Answer 3

Actually I got hold of Hobson's book and it has the expressions to second order anyway. They are: $$ P_n(\cosh x) ~ \substack{\huge\rightarrow\\\scriptstyle{n\rightarrow\infty}}~ \frac{1}{\sqrt{\pi n}}\frac{e^{(n+1/2)x}}{\sqrt{2\sinh x}}\bigg(1-\frac{2-\coth x}{8n}\bigg)\\ Q_n(\cosh x)~ \substack{\huge\rightarrow\\\scriptstyle{n\rightarrow\infty}}~ \sqrt{\frac{\pi}{n}}\frac{e^{-(n+1/2)x}}{\sqrt{2\sinh x}}\bigg(1-\frac{2+\coth x}{8n}\bigg). $$ The expression for $P_n$ agrees with Carlo's answer with the asymptotic form of the Bessel function to second order and summing up to $\nu=1$.